Tính:
\(\frac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}\)
Nhớ ghi cả cách làm
\(\frac{2^{13}+2^5}{2^{10}+2^2}=\)
\(\frac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}=\)
CẢ CÁCH LÀM NỮA NHÉ<-_->
\(\frac{21^2.14.125}{35^5.6}=\frac{\left(3.7\right)^2.2.7.5^3}{\left(5.7\right)^5.2.3}=\frac{3^2.7^3.2.5^3}{5^5.2.3.7^5}=\frac{3.}{5^2.7^2}=\frac{3}{1225}\)
b,
\(\frac{72^3.54^2}{108^4}=\frac{\left(2^3.3^2\right)^3.\left(2.3^3\right)^2}{\left(2^2.3^3\right)^4}=\frac{2^{11}.3^{12}}{2^8.3^{12}}=2^3=8\)
Ai là sky thì tick cho tui nha,ko tick ko phải là sky của sếp Tùng.
a, 105
b,8
c,8
d3
lam rui nho tik nha.khi nao ra mk lam tip cho
\(\frac{2^{13}+2^5}{2^{10}+2^2}\)
\(\frac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}\)
\(\frac{2^{13}+2^5}{2^{10}+2^2}=\frac{2^5.\left(2^8+1\right)}{2^2.\left(2^8+1\right)}=2^3=8\)
\(\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}.\left(13+65\right)}{2^8.104}=\frac{2^2.78}{104}=3\)
Rút gọn biểu thức:
\(\frac{2^{12}\cdot13+2^{12}\cdot65}{2^{10}\cdot104}+\frac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}\)
\(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
\(\Rightarrow\frac{2^{12}.\left(13+65\right)}{2^{10}.104}+\frac{3^{10}.\left(11+5\right)}{3^9.2^4}\)
\(\Rightarrow\frac{2^{12}.78}{2^{10}.104}+\frac{3^{10}.2^4}{3^9.2^4}\)
\(=\frac{2^2.3}{4}+3\)
\(=3+3=6\)
b\(^{\frac{2^{10\cdot13+2^{10}\cdot65}}{2^8\cdot104}}\)
c) \(\frac{5\cdot4^{15}-99-4\cdot3^{20}\cdot89}{5\cdot29\cdot6^{19}\cdot7\cdot2^{29}\cdot27^6}\)
b\(^{\frac{2^{10\cdot13+2^{10}\cdot65}}{2^8\cdot104}}\)
c) \(\frac{5\cdot4^{15}-99-4\cdot3^{20}\cdot89}{5\cdot29\cdot6^{19}\cdot7\cdot2^{29}\cdot27^6}\)
Hãy so sánh B và C: \(B=\frac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}\) và \(C=\frac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}\)
B=3^10.11+3^10.5/3^9.2^4
= 3^10( 11+5)/3^9.16
= 3^10.16/3^9.16
= 3^10/3^9
= 3
Vậy B = 3 (1)
C = 2^10.13+2^10.65/2^8.104
= 2^10(13+65)/2^8.2^2.26
= 2^10.78/2^10.26
= 78/26
= 3
Vậy C = 3 (2)
Từ (1) v (2) suy ra B=C
Tính gíá trị biểu thức :
A= \(\frac{3^{10}\cdot11+\left(3^2\right)^5:5}{24\cdot27}\)
B= \(\frac{2^{10}\cdot13+2^{10}\cdot65}{28\cdot104}\)
AI NHANH MK K CHO
cách làm :
câu B
ta thấy có 2 lần 210 xuất hiện trên 1 phần tử
ta gộp lại như sau :
210 x ( 13 + 65 ) cho dễ
còn câu A
ta không thể tóm gọn nên phải tính như bình thường
A = 649539 + 45 / 648
A = 1002 . 44444444444444444
hay 649584 / 648
B = 0 + 66560 / 2912
B = 22 . 8571428571
hay 66560 / 2912
Đ/s : .........
nhé !
cho mình sửa lại câu B
B = 1024 x ( 13 + 65 ) / 2912
B = 79872 / 2912
a) \(\frac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}\)
b) (1+2+3+4+....+100)x(\(1^2\) +\(2^2\) \(3^2\) +....+\(10^2\) )x(65 x 111 -13x15x37)
a.
\(\frac{2^{10}\times13+2^{10}\times65}{2^8\times104}=\frac{2^{10}\times\left(13+65\right)}{2^8\times104}=\frac{2^2\times78}{104}=\frac{4\times78}{104}=\frac{312}{104}=3\)
b.
\(\left(1+2+...+100\right)\times\left(1^2+2^2+...+10^2\right)\times\left(65\times111-15\times37\times13\right)\)
\(=\left(1+2+...+100\right)\times\left(1^2+2^2+...+10^2\right)\times\left(7215-7215\right)\)
\(=\left(1+2+...+100\right)\times\left(1^2+2^2+...+10^2\right)\times0\)
= 0
Bài 22 : a, Cho A = 4 + 22 + 23 + 24 + ... + 220
Hỏi A có chia hết cho 128 không ?
b, Tính giá trị biểu thức
\(\frac{2^{12}\cdot13+2^{12}\cdot65}{2^{10}\cdot104}+\frac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}\)
b)
\(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
\(=\frac{2^{10}\left(4.13+4.65\right)}{2^{10}.104}+\frac{3^9\left(11.3+5.3\right)}{3^9.16}\)
\(=\frac{312}{104}+\frac{48}{16}=3+3=6\)
a) \(A=4+2^2+2^3+2^4+....+2^{20}\)
\(\Rightarrow2A=2^3+2^3+2^4+.....+2^{21}\)
\(\Rightarrow2A-A=\left(2^3+2^3+2^4+....+2^{21}\right)-\left(2^2+2^3+2^4+...+2^{20}\right)\)
\(\Rightarrow A=2^3+2^{21}-\left(2^2+2^2\right)\)
\(\Rightarrow A=2^{21}\)
\(\text{Vì }2^{21}⋮2^7\Rightarrow A⋮128\)
b) \(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
\(=\frac{2^{12}\left(13+65\right)}{2^{10}.2^3.13}+\frac{3^{10}\left(11+5\right)}{3^9.2^4}\)
\(=\frac{2^{12}.78}{2^{13}.13}+\frac{3^{10}.16}{3^9.16}=\frac{6}{2}+\frac{3^{10}}{3^9}\)
\(=3+3=6\)